ASHAY DHARWADKER

We present a new proof of the famous four colour theorem using algebraic and topological methods. Recent research in physics shows that this proof directly implies the Grand Unification of the Standard Model with Quantum Gravity in its physical interpretation and conversely the existence of the standard model of particle physics shows that nature applies this proof of the four colour theorem at the most fundamental level, giving us a grand unified theory. In particular, we have shown how to use this theory to predict the Higgs Boson Mass[arXiv:0912.5189] with precision. Thus, nature itself demonstrates the logical completeness and consistency of the proof. This proof was first announced by the Canadian
Mathematical Society in 2000. The proof appears as the twelfth chapter of the text book Graph Theory published by Orient Longman and Universities Press of India in 2008. This proof has also been published in the Euroacademy Series Baltic Horizons No. 14 (111) dedicated to Fundamental Research in Mathematics in 2010. Finally, the proof features in an exquisitely illustrated edition of The Four Colour Theorem published by Amazon in 2011. The Endowed Chair of the Institute of Mathematics in recognition of this achievement was bestowed in 2012.

The famous four colour theorem seems to
have been first proposed by Möbius in 1840, later by DeMorgan and
the Guthrie brothers in 1852, and again by Cayley in 1878. The problem
of proving this theorem has a distinguished history, details of which abound
in the literature. The statement of the theorem may be introduced as follows.
In colouring a geographical map it is customary to give different colours
to any two countries that have a segment of their boundaries in common.
It has been found empirically that any map, no matter how many countries
it contains nor how they are situated, can be so coloured by using only
four different colours. The map of India requires four colours in the states
bordering Madhya Pradesh. The fact that no map was ever found whose colouring
requires more than four colours suggests the mathematical theorem.

FOUR COLOUR THEOREM.For any subdivision of the plane into
non-overlapping regions, it is always possible to mark each of the regions
with one of the numbers 0, 1, 2, 3 in such a way that no two
adjacent regions receive the same number.

STEPS OF THE PROOF: We shall outline the strategy of the new proof given
in this paper. In section I on MAP COLOURING,
we define maps on the sphere and their proper colouring. For purposes of
proper colouring it is equivalent to consider maps on the plane and furthermore,
only maps which have exactly three edges meeting at each vertex. Lemma
1 proves the six colour theorem using Euler's formula, showing that any
map on the plane may be properly coloured by using at most six colours.
We may then make the following basic definitions.

Define N to be the minimal number of colours required to properly
colour any map from the class of all maps on the plane.

Based on the definition of N, select a specific map m(N)
on the plane which requires no fewer than N colours to be properly
coloured.

Based on the definition of the map m(N), select a
proper colouring of the regions of the map m(N) using
the N colours 0, 1, ..., N-1.

The whole proof works with the fixed number N, the fixed map m(N)
and the fixed proper colouring of the regions of the map m(N).
In section II we define STEINER SYSTEMS
and prove Tits' inequality and its consequence that if a Steiner system
S(N+1,
2N, 6N) exists, then N cannot exceed 4. Now
the goal is to demonstrate the existence of such a Steiner
system. In section III we define EILENBERG
MODULES. The regions of the map m(N)
are partitioned into disjoint, nonempty equivalence classes 0, 1,
..., N-1 according to the colour they receive. This set is
given the structure of the cyclic group ZN = {0,
1,
...,
N-1} under addition modulo N. We regard
ZN
as an Eilenberg module for the symmetric group
S3 on
three letters and consider the split extension
ZN]S3
corresponding
to the trivial representation of S3. By section IV on
HALL
MATCHINGS we are able to choose a common system of coset
representatives for the left and right cosets of S3 in
the full symmetric group on |ZN]S3|
letters. For each such common representative and for each ordered pair
of elements of
S3, in section V on RIEMANN
SURFACESwe establish a certain action of the two-element
cyclic group on twelve copies of the partitioned map m(N)
by using the twenty-fourth root function of the sheets of the complex plane.
Using this action, section VI gives the details of the MAIN
CONSTRUCTION. The 6N elements of
ZN]S3
are regarded as the set of points and lemma 23 builds the blocks of 2N
points with every set of N+1 points contained in a unique block.
This constructs a Steiner system S(N+1, 2N, 6N)
which implies by Tits' inequality that N cannot exceed 4, completing
the proof. The lemmas 1-23 and theorem 24 below are written in logical
sequence. ☐

I. MAP COLOURING

A map on the sphere is a subdivision
of the surface into finitely many regions. A map is regarded as properly
coloured if each region receives a colour and no two regions having
a whole segment of their boundaries in common receive the same colour.
Since deformations of the regions and their boundary lines do not affect
the proper colouring of a map, we shall confine ourselves to maps whose
regions are bounded by simple closed polygons. For purposes of proper colouring
it is equivalent to consider maps drawn on the plane. Any map on the sphere
may be represented on the plane by boring a small hole through the interior
of one of the regions and deforming the resulting surface until it is flat.
Conversely, by a reversal of this process, any map on the plane may be
represented on the sphere. Furthermore, it suffices to consider 3-regular
maps, i.e. maps with exactly three edges meeting at each vertex, by
the following argument. Replace each vertex at which more than three edges
meet by a small circle and join the interior of each such circle to one
of the regions meeting at the vertex. A new map is obtained which is 3-regular.
If this new map can be properly coloured by using at most n colours,
then by shrinking the circles down to points, the desired colouring of
the original map using at most n colours is obtained.

Figure 1. A map that requires four colours
to be properly coloured

1. LEMMA.Any map on the sphere can be properly coloured by
using at most six colours.

PROOF: Assume that the given map is 3-regular. First show that there
must be at least one region whose boundary is a polygon with fewer than
six sides, as follows. Let E be the number of edges, V the
number of vertices, F the number of regions and Fn
the number of regions whose boundary is a polygon with n sides in
the given map. Then

F = F2+F3+F4+...

2E = 3V = 2F2+3F3+4F4+...

since a region bounded by n edges has n vertices and each
vertex belongs to three regions. By Euler's formula V-E+F
= 2,

⇒

6V-6E+6F = 12

⇒

4E-6E+6F = 12

⇒

6F-2E = 12

⇒

6(F2+F3+F4+...)-(2F2+3F3+4F4+...) = 12

⇒

4F2+3F3+2F4+F5+0+NEGATIVE TERMS = 12

.

Hence, at least one of F2, F3, F4,
F5
must be positive. Now, if a region R with fewer than six sides is
removed from the map and the resulting map coloured with six colours inductively,
there is always a colour left for R. ☐

By lemma 1, the minimal number of colours required to properly colour
any map from the class of all maps on the sphere is a well-defined natural
number. We may now make the following basic definition.

DEFINITION

Define N to be the minimal number of colours required to properly
colour any map from the class of all maps on the sphere. That is, given
any map on the sphere, no more than N colours are required to properly
colour it and there exists a map on the sphere which requires no fewer
than N colours to be properly coloured.

Based on the definition of N, select a specific map m(N)
on the sphere which requires no fewer than N colours to be properly
coloured.

Based on the definition of the map m(N), select a
proper colouring of the regions of the map m(N) using
the N colours 0, 1, ..., N-1.

The natural number N, the map m(N) and the
proper colouring of the regions of m(N) is fixed for
all future reference. By the example shown in figure 1 and lemma 1, 4 ≤ N ≤ 6.
The goal is to show that N ≤ 4.

II. STEINER SYSTEMS

A Steiner systemS(t,
k,
v)
is a set P of points together with a set
B
of blocks such that

There are v points.

Each block consists of k points.

Every set of t points is contained in a unique block.

Note that by definition t, k, v are nonnegative integers
with t ≤ k ≤ v.
Steiner systems with v = k (only one block that contains
all the points) or k = t (every k-element subset of
points is a block) are called trivial. An example of a nontrivial Steiner
system is S(5, 8, 24) due to Witt, whose blocks are known as Golay
codewords of weight eight [see
an explicit construction]. The group of automorphisms of S(5,
8, 24) (permutations of points which permute blocks) is the largest of
the Mathieu groups, M24.

2. LEMMA. [J. TITS] If there exists a nontrivial Steiner system
S(t,
k,
v)
then

v ≥ (t+1)(k-t+1).

PROOF: First show that there exists a set X0 of t+1
points that is not contained in any block, as follows. Suppose that for
every set X of t+1 points there is a block BX
that contains it. Then, this block BX must be the unique
block containing X, since X has more than t points.
Let b denote the total number of blocks. Count in two ways the number
of pairs (X, BX) where X is a set of t+1
points and BX is the unique block containing it. One
finds

(

v

t+1

)

=

b

(

k

t+1

)

.

Count in two ways the number of pairs (Y, BY)
where Y is a set of t points and BY is
the unique block containing it, by definition of a Steiner system. One
finds

(

v

t

)

=

b

(

k

t

)

.

Hence

(

v

t+1

)

(

v

t

)

=

=

b.

(

k

t+1

)

(

k

t

)

and it follows that b = 1 and k = v, contradicting
the hypothesis that the Steiner system is nontrivial. Now choose a fixed
set X0 of t+1 points that is not contained in
any block. For each set Z of t points contained in X0
there is a unique block BZ containing Z. Each
such BZ has k-t points not in X0
and any point not in X0 is contained in at most one such
BZ,
since two such blocks already have t-1 points of X0
in common. The union of the blocks BZ contains (t+1)+(t+1)(k-t)
points and this number cannot exceed the total number of points v.
☐

Recall the definition from section I that N is the minimal number
of colours required to properly colour any map from the class of all maps
on the sphere and m(N) is a specific map which requires
all of the N colours to properly colour it. The regions of the map
m(N)
have been properly coloured using the N colours 0, 1, ..., N-1.
From the map m(N) and its fixed proper colouring,
we shall construct a Steiner system S(N+1, 2N, 6N)
by defining the points and blocks in a certain way. The next lemma shows
that this construction would force N ≤ 4.

3. LEMMA.Referring to the definition ofNin
section I, if there exists a Steiner system S(N+1, 2N,
6N) then N ≤ 4.

Now, the goal is to demonstrate the existence of the Steiner system
S(N+1,
2N, 6N) based upon the definition of the map m(N).

III. EILENBERG MODULES

Let G be a group with identity
element e and let Z denote the integers. The integral
group algebra (ZG, +, ·) is a ring whose elements are
formal sums

Σ

ngg

g∈G

with g in G and ng in Z
such that ng = 0 for all but a finite number of g.
Addition and multiplication in ZG are defined by

Σ

ng g

+

Σ

mg g

=

Σ

(ng+mg)g,

g∈G

g∈G

g∈G

Σ

ng g

·

Σ

mg g

=

Σ

(

Σ

ngh-1mh

)

g.

g∈G

g∈G

g∈G

h∈G

The element n of Zis identified with the element
n·e
of ZG and the element g of
G is identified
with the element 1·g of ZG, so that Z
and G are to be regarded as subsets of ZG. The underlying
additive abelian group (ZG, +) is the direct sum of copies
of the integers Z indexed by elements of G. If Q
is a subgroup of G then ZQ is a subring of ZG
in a natural way. For each element g of G, the right multiplication
R(g):
G → G;
x → xg
and the left multiplication
L(g): G → G;
x → gx
are permutations of the set G. Denote the group of all permutations
of the set G by Sym(G). Then

R

: G → Sym(G);
g → R(g)

L-1

: G → Sym(G)
; g → L-1(g)
= L(g-1)

are embeddings of the group G in Sym(G). The images
R(G),
L-1(G)
are called the Cayley right and left regular representations of G,
respectively. The subgroup of Sym(G) generated by the set
R(G)∪L-1(G)
= {R(g), L(g-1)|g∈G}
is called the combinatorial multiplication groupMlt(G)
of
G. There is an exact sequence of groups

Δ

T

1

→

C(G)

→

G×G

→

Mlt(G)

→

1

where T(x, y) = R(x)L(y-1)
and Δc = (c, c) for an
element c of the center C(G) of G. If Q
is a subgroup of G then the relative combinatorial multiplication
groupMltG(Q)
ofQin G
is the subgroup of Mlt(G) generated by the set R(Q)∪L-1(Q)
= {R(q), L-1(q)|q∈Q}.
The orbits of the action of MltG(Q) on G
are the double cosets QgQ of the subgroup Q in G.
The stabilizer of the identity element e is the subgroup of MltG(Q)
generated by the set {T(q) = R(q)L-1(q)|q∈Q}.
A representation of the group Q is usually defined as a module,
i.e. an abelian group (M, +), for which there is a homomorphism
T:
Q → Aut(M,
+) showing how Q acts as a group of automorphisms of the module.
Another approach due to Eilenberg, views a module M for the group
Q
as follows. The set M×Q
equipped with the multiplication

(m1, q1)(m2,
q2)
= (m1 + m2T(q1),
q1q2)

becomes a group M]Q known as the split extension of
M by Q. There is an exact sequence of groups

ι

π

1

→

M

→

M]Q

→

Q

→

1

with ι: M → M]Q;
m → (m,
e)
and π: M]Q → Q;
(m, q) → q split by 0:
Q → M]Q;
q → (0,
q).
The group action
T is recovered from the split extension
M]Q
by mT(q)ι =
mιR((0,
q))L-1((0,
q))
for m in M and q in Q. In this context
we shall call M an Eilenberg module for the group Q. For
example, the trivial representation for the group Q is obtained
by defining
T: Q → Aut(M,
+);
q → 1M, the identity
automorphism of (M, +) and the corresponding split extension is
the group direct product M×Q.
The Cayley right regular representation for the group Q is obtained
by defining

T: Q → Aut(ZQ,
+); q →

(

Σ

ngg

→

Σ

nggR(q)

)

.

g∈Q

g∈Q

Here T(q) = R(q)L-1(q)
with L-1(q) acting trivially on the module elements
and R(q) acting as the usual right multiplication. The split
extension ZQ]Q has multiplication given by

(m1, q1)(m2,
q2)
= (m1+ m2R(q1),
q1q2)

for m1, m2 in ZQ and
q1,
q2
in Q.

Referring to the definition in section I, N is the minimal number
of colours required to properly colour any map from the class of all maps
on the sphere and m(N) is a specific map that requires
all of N colours to be properly coloured. Note that m(N)
has been properly coloured by using the N colours 0, 1, ..., N-1
and this proper colouring is fixed. The set of regions of m(N)
is then partitioned into subsets 0, 1, ..., N-1
where the subset m consists of all the regions which receive
the colour m. Note that the subsets 0, 1, ..., N-1
are each nonempty (since m(N) requires all of the
N
colours to be properly coloured) and form a partition of the set of regions
of m(N) (by virtue of proper colouring). Identify
the set {0, 1, ..., N-1} with the underlying
set of the N-element cyclic group ZN under
addition modulo N. Let S3 denote the symmetric
group on three letters, identified with the dihedral group of order six
generated by ρ, σ
where
|ρ| = 3 and |σ| =
2.

4. LEMMA. (ZN, +) is an Eilenberg
module for the groupS3with the trivial homomorphism

T1: S3 → Aut(ZN,
+); α → 1ZN

where1ZN
denotes the identity automorphism ofZN.
The
corresponding split extension ZN]S3 has
multiplication given by

(m1, α1)·(m2,
α2)
= (m1+
m2, α1α2)

and is a group isomorphic to the direct productZN×S3.

PROOF: Follows from definition. ☐

Referring to section II, the goal is to construct a Steiner system S(N+1,
2N, 6N). We shall take the point set of the Steiner system
to be the underlying set of the split extension ZN]S3.
The following lemma is used in section V.

5. LEMMA.Let (Z(ZN]S3),
+)
and (ZS3, +) denote the underlying
additive groups of the integral group algebrasZ(ZN]S3)
and
ZS3,
respectively.
Then
(Z(ZN]S3), +)
is
an Eilenberg module for the group (ZS3, +)
with
the trivial homomorphism

T2:(ZS3, +) → Aut(Z(ZN]S3),
+);

Σ

nαα → 1Z(ZN]S3)

α∈S3

where 1Z(ZN]S3)
denotes
the identity automorphism of (Z(ZN]S3),
+).
The corresponding split extensionZ(ZN]S3)]ZS3
has multiplication given by

(

Σ

n(m, β)(m,
β)

,

Σ

nαα

)

(

Σ

n'(m, β)(m,
β)

,

Σ

n'αα

)

(m, β)∈ZN]S3

α∈S3

(m, β)∈ZN]S3

α∈S3

=

(

Σ

(n(m, β)+n'(m,
β))(m,
β)

,

Σ

(nα+n'α)α

)

(m, β)∈ZN]S3

α∈S3

and is a group isomorphic to the direct product (Z(ZN]S3)×ZS3,
+).

PROOF: Follows from definition. ☐

IV. HALL MATCHINGS

Let Γ be a
bipartite graph with vertex set V = X∪Y
and edge set E (every edge has one end in X and the other
end in Y). A matching from X to Y in Γ
is a subset M of E such that no vertex is incident with more
than one edge in M. A matching M from X to Y in Γ
is called complete if every vertex in X is incident with
an edge in M. If A is a subset of V then let adj(A)
denote the set of all vertices adjacent to a vertex in A.

PROOF: A matching from X to Y in Γ
with |M| = 1 always exists by choosing a single edge in E.
Let M be a matching from X to Y in Γ
with m edges, m < |X|.
Let x0∈X such
that x0 is not incident with any edge in M. Since
|adj({x0})| ≥ 1,
there is a vertex y1 adjacent to x0
by an edge in E\M. If y1 is not incident with
an edge in M, then stop. Otherwise, let x1 be
the other end of such an edge. If x0, x1,
..., xk and y1, ..., yk
have been chosen, then since |adj({x0, x1,
..., xk})| ≥ k+1,
there is a vertex yk+1, distinct from y1,
..., yk, that is adjacent to at least one vertex in {x0,
x1,
..., xk}. If yk+1 is not
incident with an edge in M, then stop. Otherwise, let xk+1
be the other end of such an edge. This process must terminate with some
vertex, say yk+1. Now build a simple path
from yk+1 to x0 as follows.
Start with yk+1 and the edge in E\M
joining it to, say xi1
with i1 < k+1. Then
add the edge in M from xi1
to yi1. By
construction, yi1
is joined by an edge in E\M to some xi2
with i2 < i1.
Continue adding edges in this way until x0 is reached.
One obtains a path yk+1, xi1,
yi1,
xi2,
yi2,
..., xir,
yir,
x0
of odd length 2r+1 with the r+1 edges {yk+1,
xi1},
{yi1,
xi2},
..., {yir, x0}
in E\M and the r edges {xi1,
yi1},
..., {xir, yir}
in M. Define

Then M' is a matching from X to Y in Γ,
with |M'| = |M|-r+r+1 = |M|+1. Repeating
this process a finite number of times must yield a complete matching from
X
to Y in Γ. ☐

7. LEMMA.Referring to section III, letSym(ZN]S3)
denote
the group of all permutations of the underlying set of the split extension
ZN]S3
of
lemma 4. ThenS3
embeds inSym(ZN]S3)
via
the Cayley right regular representation.

PROOF: Note that S3 = {(0, α)|α∈S3}
is a subgroup of ZN]S3. Since
S3
embeds in Sym(S3) via the Cayley right regular
representation
α → R(α)
and Sym(S3) is a subgroup of Sym(ZN]S3),
the lemma follows. ☐

8. LEMMA.By lemma 7, regardS3
as a subgroup ofSym(ZN]S3).
There
exists a common system of coset representatives φ1,
..., φk such that {φ1S3,
..., φkS3}
is
the family of left cosets ofS3inSym(ZN]S3)
and
{S3φ1, ..., S3φk}
is
the family of right cosets ofS3inSym(ZN]S3).

PROOF: By Lagrange's theorem the left cosets of S3
partition Sym(ZN]S3)
into k = [Sym(ZN]S3):S3]
disjoint nonempty equivalence classes of size |S3| =
6. The same is true of the right cosets. Define a bipartite graph Γ
with vertices X∪Y where
X
= {ψ1S3, ..., ψkS3}
is the family of left cosets of S3 in Sym(ZN]S3)
and Y = {S3ψ'1,
..., S3ψ'k} is
the family of right cosets of S3 in Sym(ZN]S3)
with an edge {ψiS3,
S3ψ'j}
if and only if ψiS3
and S3ψ'j
have nonempty intersection. Note that we can select representatives of
the left cosets that belong to distinct right cosets [see
a proof of this fact]. For any subset A = {ψi1S3,
..., ψirS3}
of X, one has ψi1∈ψi1S3,
..., ψir∈ψirS3
and there exist distinct j1, ..., jr
such that ψi1∈S3ψ'j1,
..., ψir∈S3ψ'jr.
Hence, in the graph Γ, |adj(A)| ≥ |A|.
Hall's hypothesis of lemma 6 is satisfied and there exists a complete matching
from X to Y in Γ. This is precisely
the statement that a common system of coset representatives φ1,
..., φk exists. ☐

and the whole w-plane except for the positive real axis. The
image of each sector is obtained by performing a cut along the positive
real axis; this cut has an upper and a lower edge. Corresponding to the
n
sectors in the z-plane, take n identical copies of the w-plane
with the cut. These will be the sheets of the Riemann surface and
are distinguished by a label k which serves to identify the corresponding
sector. For k = 1, ..., n-1 attach the lower edge of the
sheet labeled k with the upper edge of the sheet labeled k+1.
To complete the cycle, attach the lower edge of the sheet labeled n
to the upper edge of the sheet labeled 1. In a physical sense, this is
not possible without self-intersection but the idealized model shall be
free of this discrepancy. The result of the construction is a
Riemann
surface whose points are in one-to-one correspondence with the points
of the z-plane [see
a geometric model]. This correspondence is continuous in the following
sense. When z moves in its plane, the corresponding point w
is free to move on the Riemann surface. The point w = 0
connects all the sheets and is called the branch point. A curve
must wind n times around the branch point before it closes. Now
consider the n-valued relation

z = n√w.

To each w ≠ 0, there correspond
n
values of z. If the w-plane is replaced by the Riemann surface
just constructed, then each complex number w ≠ 0
is represented by n points of the Riemann surface at superposed
positions. Let the point on the uppermost sheet represent the principal
value and the other n-1 points represent the other values. Then
z
= n√w becomes
a single-valued, continuous, one-to-one correspondence of the points of
the Riemann surface with the points of the z-plane. Now recall the
definition of the map m(N) from section I. The map
m(N)
is on the sphere. Pick a region and deform the sphere so that both 0 and
∞
are two distinct points inside this region when the sphere is regarded
as the extended complex plane. Using the stereographic projection one obtains
the map m(N) on the complex plane C
with the region containing 0 and ∞ forming
a ''sea'' surrounding the other regions which form an ''island''. Put this
copy of C on each sheet of the Riemann surface corresponding
to w = zn. The branch point lies in the ''sea''.
The inverse function z = n√w
results in n copies of the map m(N) on the
z-plane
in
the sectors

{z|(k-1)2π/n < arg
z < k2π/n}
(k = 1, ..., n).

The origin of the z-plane lies in the n ''seas''.

z-plane

w = z4 Riemann surface

Figure 2. An example with n = 4

Referring to section III, the full symmetric group Sym(ZN]S3)
acts faithfully on the set ZN]S3.
The action of an element ψ of Sym(ZN]S3)
on an element (m, α) of ZN]S3
will be written as (m, α)ψ.
This action extends to the integral group algebra Z(ZN]S3)
by linearity

(

Σ

n(m, α)(m,
α)

)

ψ

=

Σ

n(m, α)((m,
α)ψ)

.

(m, α)∈ZN]S3

(m, α)∈ZN]S3

Referring to lemma 8, fix a common coset representative φi
of S3 in Sym(ZN]S3)
and fix a pair (β, γ)∈S3×S3
=
Mlt(S3).
There are two cases depending on whether
β =
γ
or whether
β ≠ γ.

CASE 1. Suppose β ≠ γ. Consider
the composition of the functions

C → C; z → t
= z2 and C → C;
t → w
= t12.

The composite is given by the assignment

z → t = z2 → w
= t12 = z24.

There are twenty-four superposed copies of the map m(N)
on the w-Riemann surface corresponding to the sectors

{z|(k-1)2π/24 < arg
z < k2π/24}
(k = 1, ..., 24)

on the z-plane. These are divided into two sets. The first set
consists of twelve superposed copies of the map m(N)
corresponding to the sectors

{z|(k-1)2π/24 < arg
z < k2π/24}
(k = 1, ..., 12)

of the upper half of the z-plane which comprise the upper sheet
of the t-Riemann surface. The second set consists of twelve superposed
copies of the map m(N) corresponding to the sectors

{z|(k-1)2π/24 < arg
z < k2π/24}
(k = 13, ..., 24)

of the lower half of the z-plane which comprise the lower sheet
of the t-Riemann surface.

Figure 3. Sheets of the t-Riemann surface

Label the twelve sectors of the upper sheet of the t-Riemann
surface by elements of Z(ZN]S3)]ZS3
as shown:

Figure 4. Upper sheet of the t-Riemann
surface

Label the twelve sectors of the lower sheet of the t-Riemann
surface by elements of Z(ZN]S3)]ZS3
as shown:

Figure 5. Lower sheet of the t-Riemann
surface

Referring to section II, the regions of the map m(N)
have been partitioned into disjoint, nonempty equivalence classes 0,
1,
..., N-1 and this set of equivalence classes forms the underlying
set of the cyclic group ZN. Hence there are twelve
copies of ZN on the upper sheet and twelve copies
of ZN on the lower sheet of the t-Riemann
surface. The copies of ZN are indexed by the elements
of Z(ZN]S3)]ZS3
which label the sectors on a particular sheet. The branch point of the
t-Riemann
surface is labeled by the element (0, β+γ)
of Z(ZN]S3)]ZS3
where 0 denotes the zero element of Z(ZN]S3).

9. LEMMA.Referring to lemma 8, fix a common representative
φi
of
the left and right cosets of S3inSym(ZN]S3).
Fix
a pair (β, γ)∈S3×S3
with
β ≠ γ.
Referring
to lemma 5,
define a subsetT(β,
γ)
of
Z(ZN]S3)]ZS3
as
follows:

T(β, γ)
=

{((m, α),
β+γ)|(m,
α)∈ZN]S3}

∪

{(0, β+γ)}

∪

{(-(m, α),
β+γ)|(m,
α)∈ZN]S3}

.

Referring to the preceding discussion, consider the composite function

z → t = z2 → w
= t12 = z24

of the complex z-plane to thew-Riemann surface.
There
is a copy of the set T(β,
γ)
on
the upper sheet and a copy of the setT(β,
γ)
on
the lower sheet of thet-Riemann surface according to the
labels of the sectors in figures 4 and 5 with the branch
point labeled by the element (0, β+γ)
of
both copies. The rotation of thez-plane by π
radians
induces a permutation

p: T(β, γ) → T(β,
γ)

given by

(-(m, α)R(γ)φi,
β+γ)p

=

((m, α)φiR(γ),
β+γ)

(0, β+γ)p

=

(0, β+γ)

((m, α)φiR(β),
β+γ)p

=

(-(m, α)R(β)φi,
β+γ)

for all (m, α)∈ZN]S3,
such
that each point of the copy ofT(β,
γ)
on
the upper sheet moves continuously along a circular curve that winds exactly
once around the branch point, to the point superposed directly below it
on the copy ofT(β,
γ)
on
the lower sheet of thet-Riemann surface.

PROOF: T(β, γ)
is seen to be a well-defined subset of Z(ZN]S3)]ZS3
by setting the appropriate coefficients to zero in a typical element as
described in lemma 5. Each of R(γ)φi,
φiR(γ),
φiR(β),
R(β)φi
are permutations of the set ZN]S3
and the rotation of the z-plane by π
radians clearly induces a permutation p of the set T(β,
γ)
as described. ☐

PROOF: If p = 1, then (-(0,1)R(γ)φi,
β+γ)
= (-(0,1)R(γ)φi,
β+γ)p
= ((0,1)φiR(γ),
β+γ)
which implies that -(0,1)R(γ)φi=
(0,1)φiR(γ)
in Z(ZN]S3). This
is impossible since 1 ≠ -1 in Z.
Hence p ≠ 1. Since the full permutation
group Sym(T(β, γ))
acts faithfully on T(β, γ),
so does its subgroup < p > .
☐

11. LEMMA.Referring to lemma 9 and lemma 10, let
1:
C → C;
z → z
denote the identity and π: C → C;
z → -z
denote the rotation through an angle of π
radians of the z-plane. Then the two-element cyclic group
{1,
π}
acts faithfully on the set < p >
as
follows:pn1 = pn and pnπ
= p1-n,
for all n in Z.

PROOF: The set {1, π} forms a two-element
cyclic group < π > under function composition.
To show that {1, π} acts on < p >
as defined, observe that (pnπ)π
= (p1-n)π = p1-(1-n)
= p1-1+n = pn = pn1
= pn(ππ), for all n
in Z. To show that the action is faithful, let θ∈{1,
π}.
If θ belongs to the kernel of the action then
pnθ
= pn for all n in Zso that pθ
= p which implies that θ = 1, since p ≠ 1
by lemma 10. ☐

12. LEMMA.Putting together lemma 9, lemma 10 and
lemma 11, there is a well-defined action of the two-element cyclic
group {1, π} on the setT(β,
γ)
given
by

((m, α)φiR(γ),
β+γ)1

=

((m, α)φiR(γ),
β+γ)

(0, β+γ)1

=

(0, β+γ)

(-(m, α)R(β)φi,
β+γ)1

=

(-(m, α)R(β)φi,
β+γ)

and

((m, α)φiR(γ),
β+γ)π

=

(-(m, α)R(γ)φi,
β+γ)

(0, β+γ)π

=

(0, β+γ)

(-(m, α)R(β)φi,
β+γ)π

=

((m,α)φiR(β),
β+γ)

for all (m, α) in
ZN]S3.
This
action is faithful.

PROOF: For each x∈T(β,
γ),
let Orb(x) = {xpn|n∈Z}
denote the orbit of x under < p > .
The collection {Orb(x)|x∈T(β,
γ)}
forms a partition of the set T(β,
γ)
as follows. Let x, y∈T(β,
γ).
If z∈Orb(x)∩Orb(y)
then z = xpn = ypm for some
m,
n∈Z.
This implies
xpn-m= y ⇒
y∈Orb(x)
⇒
Orb(y)⊆Orb(x)
and ypm-n= x ⇒
x∈Orb(y)
⇒
Orb(x)⊆Orb(y).
Hence Orb(x) = Orb(y). Also each x∈T(β,
γ)
belongs to an orbit, namely x∈Orb(x).
Hence the orbits are disjoint, nonempty and their union is all of the set
T(β,
γ).
For each fixed x∈T(β,
γ)
define

orbit by orbit. Then, since {Orb(x)|x∈T(β,
γ)}
forms a partition of T(β,
γ),
π
is a well-defined permutation of T(β,
γ)
with π2
= 1, the identity permutation
of T(β,
γ).
Hence, using the definition of p in lemma 9, we obtain an action
of the two-element cyclic group {1, π} on T(β,
γ)
as follows. For all (m,
α) in
ZN]S3
define

((-(m, α)R(γ)φi,
β+γ)p)1

=

(-(m, α)R(γ)φi,
β+γ)p

((0, β+γ)p)1

=

(0, β+γ)p

(((m, α)φiR(β),
β+γ)p)1

=

((m, α)φiR(β),
β+γ)p

and

((-(m, α)R(γ)φi,
β+γ)p)π

=

(-(m, α)R(γ)φi,
β+γ)

((0, β+γ)p)π

=

(0, β+γ)

(((m, α)φiR(β),
β+γ)p)π

=

((m, α)φiR(β),
β+γ).

Now, using the definition of p in lemma 9, the action of {1,
π}
on T(β,
γ)
may be rewritten

((m, α)φiR(γ),
β+γ)1

=

((m, α)φiR(γ),
β+γ)

(0, β+γ)1

=

(0, β+γ)

(-(m, α)R(β)φi,
β+γ)1

=

(-(m, α)R(β)φi,
β+γ)

and

((m, α)φiR(γ),
β+γ)π

=

(-(m, α)R(γ)φi,
β+γ)

(0, β+γ)π

=

(0, β+γ)

(-(m, α)R(β)φi,
β+γ)π

=

((m, α)φiR(β),
β+γ)

for all (m, α) inZN]S3,
as in the statement of this lemma. To verify that the action of {1, π}
on T(β, γ)
is faithful, note that

1: T(β, γ) → T(β,
γ);
x → x

π: T(β,
γ) → T(β,
γ);
x → xπ

are permutations of the set T(β,
γ).
If θ∈{1,
π}and
θ
belongs to the kernel of the action then xθ
= x for all x∈T(β,γ).
Then θ = 1, since π
moves ((0, 1)φiR(γ),
β+γ)
to (-(0, 1)R(γ)φi,
β+γ)
which are distinct elements of Z(ZN]S3)]ZS3.
☐

CASE 2. Suppose β = γ.
Note that in the labeling of the sectors of the sheets of the t-Riemann
surface in figures 4 and 5, R(β) = R(γ)
and β+γ = 2β
in the group algebra ZS3.

13. LEMMA.Referring to lemma 8, fix a common representative
φi
of
the left and right cosets of S3inSym(ZN]S3).
Fix
a pair (β, β)∈S3×S3.
Referring
to lemma 5, define a subsetT(β,
β)
of
Z(ZN]S3)]ZS3
as
follows:

T(β, β)
=

{((m, α),
2β)|(m, α)∈ZN]S3}

∪

{(0, 2β)}

∪

{(-(m, α),
2β)|(m, α)∈ZN]S3}

.

Referring to the preceding discussion, consider the composite function

z → t = z2 → w
= t12 = z24

of the complex z-plane to thew-Riemann surface.
There
is a copy of the set T(β,
β)
on
the upper sheet and a copy of the setT(β,
β)
on
the lower sheet of thet-Riemann surface according to the
labels of the sectors in figures 4 and 5. Note that in this
caseR(β) = R(γ)
and
β+γ
= 2β
with the branch point labeled by the
element (0, 2β)
of both copies. The
rotation of thez-plane by π
radians
induces a permutation

p: T(β, β) → T(β,
β)

given by

(-(m, α)R(β)φi,
2β)p

=

((m, α)φiR(β),
2β)

(0, 2β)p

=

(0, 2β)

((m, α)φiR(β),
2β)p

=

(-(m, α)R(β)φi,
2β)

for all (m, α)∈ZN]S3,
such
that each point of the copy ofT(β,
β)
on
the upper sheet moves continuously along a circular curve that winds exactly
once around the branch point, to the point superposed directly below it
on the copy ofT(β,
β)
on
the lower sheet of thet-Riemann surface. Then p
= p-1 so that < p >
= {1,
p}
is a two-element cyclic subgroup of the full permutation
group Sym(T(β, β))
and
< p >
acts
faithfully on the setT(β, β).

PROOF: As in the proof of lemma 9, T(β,
β)
is seen to be a well-defined subset of Z(ZN]S3)]ZS3
by setting the appropriate coefficients to zero in a typical element as
described in lemma 5. Both φiR(β),
R(β)φi
are permutations of the set ZN]S3
and the rotation of the z-plane by π
radians clearly induces a permutation p of the set T(β,
β)
as described. Furthermore, it is clear from the definition that p
= p-1 by chasing elements of T(β,
β).
Then < p >
= {1, p} as a subgroup of Sym(T(β,
β))
and < p >
acts faithfully on the set T(β,
β).
☐

14. LEMMA.Referring to lemma 13, let 1: C → C;
z → z
denote the identity and π: C → C;
z → -z
denote the rotation through an angle of π radians
of the z-plane. Then there is a well-defined action of the
two-element cyclic group {1, π}
on the
setT(β,
β)
given
by

RÉSUMÉ. Let us review
the final goal. Recall the definition made in section I. We have defined
N
to be the minimal number of colours required to properly colour any map
from the class of all maps on the sphere. We know that 4 ≤ N ≤ 6.
We have chosen a specific map m(N) on the sphere which
requires all of the N colours 0, 1, ..., N-1 to properly
colour it. The map m(N) has been properly coloured
and the regions of m(N) partitioned into disjoint,
nonempty equivalence classes 0, 1, ..., N-1
according to the colour they receive. The set {0, 1, ...,
N-1}
is endowed with the structure of the cyclic group ZN
under addition modulo N. In section III we have built the split
extension
ZN]S3. The underlying
set ZN]S3 of cardinality 6N
is taken to be the point set of a Steiner system S(N+1, 2N,
6N) which will be constructed in this section. We are required to
define the blocks of size 2N and show that every set of N+1
points is contained in a unique block. Once this goal is achieved, lemma
3 shows that N ≤ 4.

15. LEMMA.LetZN]S3
denote
the split extension defined in lemma 4 andSym(ZN]S3)
denote
the full permutation group on the setZN]S3.
Define

μ: Sym(ZN]S3) → Sym(ZN]S3)

by

ψ = R(γ)φi → φiR(γ)
= ψμ.

Then μ is a bijection of the set
Sym(ZN]S3) with itself.

PROOF: Referring to lemma 8, μ is well-defined
since each ψ∈Sym(ZN]S3)
may be written uniquely as ψ = R(γ)φi
for some γ∈S3 and some
φi.
Then μ is a surjection because for any ψ∈Sym(ZN]S3)
one may also write ψ = φiR(γ)
uniquely for some γ∈S3
and some φi, whence R(γ)φi → φiR(γ)
= ψ. Since Sym(ZN]S3)
is a finite set, μ must be a bijection by counting.
☐

16. LEMMA.Define the set G as follows:

G

=

{

(

ψ

ψμ

)

|

ψ∈Sym(ZN]S3)

}

=

{

(

R(γ)φi

φiR(γ)

)

|

γ∈S3

i = 1, ..., k

}

.

Define multiplication in G as follows:

(

ψ1

ψ1μ

)

(

ψ2

ψ2μ

)

=

(

(ψ1ψ2)

(ψ1ψ2)μ

)

i.e.

(

R(γ1)φi1

φi1R(γ1)

)

(

R(γ2)φi2

φi2R(γ2)

)

=

(

R(γ3)φi3

φi3R(γ3)

)

whereR(γ3)φi3
is the unique expression forR(γ1)φi1R(γ2)φi2
according to the right coset decomposition ofS3 in
Sym(ZN]S3).
Then G is
a group.

PROOF: Referring to lemma 8 and lemma 15, the set G is well-defined
by the decomposition of Sym(ZN]S3)
into the left and right cosets of S3 by the φi.
Define

μ': Sym(ZN]S3) → G
; ψ →

(

ψ

ψμ

)

.

Then μ' is a well-defined bijection of the
set Sym(ZN]S3) with G
since μ is a bijection by lemma 15. The definition
of multiplication in G mirrors the multiplication in Sym(ZN]S3)
via
μ' and is designed to make G a group
and μ' an isomorphism. ☐

17. LEMMA.Consider the setZN]S3
and
let

(

ψ

ψμ

)

=

(

R(γ)φi

φiR(γ)

)

∈G.

Define

↑

(

ψ

ψμ

)

: ZN]S3 → ZN]S3
by

(m, α) → (m,
α)

↑

(

ψ

ψμ

)

=

(m, α)

↑

(

R(γ)φi

φiR(γ)

)

= (m, α)R(γ)φi.

Define

↓

(

ψ

ψμ

)

: ZN]S3 → ZN]S3
by

(m, α) → (m,
α)

↓

(

ψ

ψμ

)

=

(m, α)

↓

(

R(γ)φi

φiR(γ)

)

= (m, α)φiR(γ).

Then

(m, α)

↑

(

ψ

ψμ

)

=

(m, α)

↓

(

ψ

ψμ

)

for all (m, α)∈ZN]S3.

Both ↑ and ↓
are
well-defined, faithful and |ZN]S3|-transitive
right actions of the groupGon the setZN]S3.

PROOF: Referring to lemma 12 and lemma 14, put β
= 1. Working in the set T(1, γ),
for each (m, α)∈ZN]S3
we have

(

(m, α)

↓

(

ψ

ψμ

)

,

1+γ

)

=

(

(m, α)

↓

(

R(γ)φi

φiR(γ)

)

,

1+γ

)

=

(

(m, α)

φiR(γ)

,

1+γ

)

=

(

-(m, α)

R(γ)φi

,

1+γ

)

π

=

(

-(m, αγ)

φi

,

1+γ

)

π

=

(

-(m, αγ)

R(1)φi

,

1+γ

)

π

=

(

(m, αγ)

φiR(1)

,

1+γ

)

=

(

(m, αγ)

φi

,

1+γ

)

=

(

(m, α)

R(γ)φi

,

1+γ

)

=

(

(m, α)

↑

(

R(γ)φi

φiR(γ)

)

,

1+γ

)

=

(

(m, α)

↑

(

ψ

ψμ

)

,

1+γ

)

using the action of the two-element cyclic group {1, π}
on the set T(1, γ) according
to lemma 12 and lemma 14. Hence

(m, α)

↑

(

ψ

ψμ

)

=

(m, α)

↓

(

ψ

ψμ

)

for all (m, α)∈ZN]S3.

Since the action ↑ is the usual action
of Sym(ZN]S3) on the
set
ZN]S3, it is faithful and
|ZN]S3|-transitive. By the last
equality, so is the ↓ action. ☐

PROOF: Note that if ψ = R(γ)φi=
1 then φi=
R(γ)-1
so that ψμ= φiR(γ)
= R(γ)-1R(γ)
= 1. Then

(

1

1μ

)

∈Hr, s

since

(mi, αi)

↑

(

1

1μ

)

= (mi, αi)1
= (mi, αi)

for i = 1, ..., r and

(nj, βj)

↓

(

1

1μ

)

= (nj, βj)1μ
= (nj, βj)1
= (nj, βj)

for j = 1, ..., s. If

(

ψ1

ψ1μ

)

and

(

ψ2

ψ2μ

)

∈Hr, s

then

(mi, αi)

↑

(

(

ψ1

ψ1μ

)

(

ψ2

ψ2μ

)

)

= (mi, αi)

↑

(

ψ1

ψ1μ

)

↑

(

ψ2

ψ2μ

)

= (mi, αi)

↑

(

ψ2

ψ2μ

)

= (mi, αi)
for i = 1, ..., r

and

(nj, βj)

↓

(

(

ψ1

ψ1μ

)

(

ψ2

ψ2μ

)

)

= (nj, βj)

↓

(

ψ1

ψ1μ

)

↓

(

ψ2

ψ2μ

)

= (nj, βj)

↓

(

ψ2

ψ2μ

)

= (nj, βj)
for j = 1, ..., s.

Hence

(

ψ1

ψ1μ

)

(

ψ2

ψ2μ

)

∈Hr, s

Since G is finite, Hr, s is a subgroup of G.
☐

Note that ZN is embedded as the subgroup {(m,
1)|m∈ZN}
in ZN]S3 and S3
is embedded as the subgroup {(0, α)|α∈S3}
in ZN]S3. Since ZN]S3
= ZN×S3
is the direct product of groups by lemma 4, both ZN
and S3 are normal subgroups. Recall the notation

S3 = < σ, ρ >
= {1, ρ, ρ2,
σ,
σρ,
σρ2}.

19. LEMMA.Define

H

=

{

(

ψ

ψμ

)

∈G

|

(m, 1)

↑

(

ψ

ψμ

)

= (m, 1)
for all m∈ZN
and

(0, σ)

↓

(

ψ

ψμ

)

= (0, σ)

}

.

Then given

(

ψ

ψμ

)

∈H

either

(m, α)

↓

(

ψ

ψμ

)

= (m, α) for all
(m, α)∈ZN]S3

or

(m, α)

↓

(

ψ

ψμ

)

= (m, ασ)
for
all (m, α)∈ZN]S3.

PROOF: H is a well-defined subgroup of G according to
lemma 18. Let

(

ψ

ψμ

)

=

(

R(γ)φi

φiR(γ)

)

∈H

and (m, α)∈ZN]S3
be given. Referring to lemmas 12 and 14, put β
= γ-1αγ.
Working in the set T(β, γ)
we have

(

(m, α)

↓

(

ψ

ψμ

)

,

β+γ

)

=

(

(m, α)

↓

(

R(γ)φi

φiR(γ)

)

,

β+γ

)

=

(

(m, α)

φiR(γ)

,

β+γ

)

=

(

-(m, α)

R(γ)φi

,

β+γ

)

π

=

(

-(m, αγ)

φi

,

β+γ

)

π

=

(

-(m, γβ)

φi

,

β+γ

)

π

=

(

-(m, γ)

R(β)φi

,

β+γ

)

π

=

(

(m, γ)

φiR(β)

,

β+γ

)

=

(

(m, 1)

R(γ)φiR(β)

,

β+γ

)

=

(

(m, 1)

R(β)

,

β+γ

)

=

(

(m, β)

,

β+γ

)

using the definition of H and the action of the two-element cyclic group
{1,π} on the set T(β,γ).
Hence

(m, α)

↓

(

ψ

ψμ

)

= (m, β) = (m,
γ-1αγ)
= (m, αγ).

Now since

(0, σ) = (0, σ)

↓

(

ψ

ψμ

)

= (0, σγ)

by hypothesis, we have σ = σγ.
Hence γσ = σγ so
that either γ = 1 or γ =
σ.
☐

20. LEMMA.Let H be the subgroup of G defined in lemma 19.
Then H is a nontrivial group of involutions of the set ZN]S3.
In particular, every nontrivial element of H is of order 2.

PROOF: Define

ψ: ZN]S3 → ZN]S3;
(m, α) → (m,
ασ).

Then

(m, 1)

↑

(

ψ

ψμ

)

= (m, 1)ψ = (m,
1σ) = (m, 1) for all
m∈ZN

and

(0, σ)

↓

(

ψ

ψμ

)

=

(0, σ)

↑

(

ψ

ψμ

)

= (0, σ)ψ
= (0, σσ) = (0, σ).

Now ψ ≠ 1, so

(

1

1μ

)

≠

(

ψ

ψμ

)

∈H,

hence H is nontrivial. To show that each nontrivial element of
H
is of order 2, let

(

ψ

ψμ

)

=

(

R(γ)φi

φiR(γ)

)

∈H.

Then by the proof of lemma 19, γ = 1 or γ
=
σ.
In particular γ2 = 1. Hence, for
any (m, α)∈ZN]S3

(m, α)

↓

(

R(γ)φi

φiR(γ)

)

2

= (m, α)

↓

(

R(γ)φi

φiR(γ)

)

↓

(

R(γ)φi

φiR(γ)

)

= (m, αγ)

↓

(

R(γ)φi

φiR(γ)

)

= (m, (αγ)γ)
= (m, α). Since the ↓
action of G on the set ZN]S3
is faithful,

(

R(γ)φi

φiR(γ)

)

2

=

(

1

1μ

)

the identity element of G. ☐

21. LEMMA.Denote the right cosets of ZN
in ZN]S3by

ZN, ZNρ,
ZNρ2,
ZNσ,
ZNσρ,
ZNσρ2.

Define

Fix↓(H) =

{

(m, α)∈ZN]S3

|

(m, α)

↓

(

ψ

ψμ

)

=

(m, α)

for all

(

ψ

ψμ

)

∈H

}

.

Then Fix↓(H) = {(m,
α)∈ZN]S3|α
= 1
or α = σ}.
The
↓
action of a nontrivial element of H transposes the coset ZNρ
with
the coset ZNρ2
and
transposes the coset ZNσρ
with
the coset ZNσρ2.

PROOF: By lemmas 19 and 20, the elements

(

ψ

ψμ

)

∈H

are of two kinds:

(i)

(m, α)

↓

(

ψ

ψμ

)

= (m, α) for all (m,
α)∈ZN]S3

in which case

(

ψ

ψμ

)

=

(

1

1μ

)

the identity element of H, and

(ii)

(m, α)

↓

(

ψ

ψμ

)

= (m, ασ)
for all (m, α)∈ZN]S3

in which case

(

1

1μ

)

≠

(

ψ

ψμ

)

is an element of order 2 in H. In the second case, compute according
to the cosets of ZN in ZN]S3:

(m, 1)

↓

(

ψ

ψμ

)

= (m, 1σ)
= (m, 1) for all m∈ZN

(m, σ)

↓

(

ψ

ψμ

)

= (m, σσ)
= (m,
σ) for all m∈ZN

(m, ρ)

↓

(

ψ

ψμ

)

= (m, ρσ)
= (m,
ρ2) for all m∈ZN

(m, ρ2)

↓

(

ψ

ψμ

)

= (m, (ρ2)σ)
= (m,
ρ) for all m∈ZN

(m, σρ)

↓

(

ψ

ψμ

)

= (m, (σρ)σ)
= (m,
σρ2) for all
m∈ZN

(m, σρ2)

↓

(

ψ

ψμ

)

= (m, (σρ2)σ)
= (m,
σρ) for all m∈ZN

and the lemma follows. ☐

22. LEMMA.Let NormG(H) denote the
normalizer of HinG. The action ↓
of
GonZN]S3
restricts
to an action ↓ of NormG(H)
on
Fix↓(H)
which is (|ZN|+1)-transitive.

PROOF: Let

(

ψ

ψμ

)

∈G.

First show that

Fix↓

(

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

)

= Fix↓(H)

↓

(

ψ

ψμ

)

as follows:

(m, α)∈Fix↓

(

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

)

⇔ (m, α)↓

(

(

ψ

ψμ

)

-1

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

)

= (m, α) for all

(

ψ*

ψ*μ

)

∈H

⇔ (m, α)↓

(

(

ψ

ψμ

)

-1

(

ψ*

ψ*μ

)

)

= (m, α)↓

(

ψ

ψμ

)

-1

for all

(

ψ*

ψ*μ

)

∈H

⇔ (m, α)↓

(

ψ

ψμ

)

-1

↓

(

ψ*

ψ*μ

)

= (m, α)↓

(

ψ

ψμ

)

-1

for all

(

ψ*

ψ*μ

)

∈H

⇔ (m, α)↓

(

ψ

ψμ

)

-1

∈Fix↓(H)

⇔ (m, α)∈Fix↓(H)↓

(

ψ

ψμ

)

.

Now let

(

ψ

ψμ

)

∈NormG(H)
=

{

(

ψ

ψμ

)

∈G

|

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

= H

}

.

Then

Fix↓(H) = Fix↓

(

(

ψ

ψμ

)

-1

Η

(

ψ

ψμ

)

)

= Fix↓(H)↓

(

ψ

ψμ

)

showing that the action restricts to an action of NormG(H)
on Fix↓(H). Now to show that
the action ↓ of NormG(H)
on Fix↓(H) = ZN∪ZNσ
is
(|ZN|+1)-transitive, label the elements of
ZN
as (m1, α1),
..., (mN, αN)
and label (0, σ) = (mN+1,
αN+1).
Let (m*1,
α*1),
..., (m*N+1,
α*N+1)
be any |ZN|+1 distinct points of Fix↓(H).
It is enough to show that there exists

(

ψ

ψμ

)

∈NormG(H)

such that

(m*i, α*i)

↓

(

ψ

ψμ

)

= (mi, αi)
for i = 1, ..., N+1.

Now there exists

(

ψ

ψμ

)

∈G

such that

(m*i, α*i)

↓

(

ψ

ψμ

)

= (mi, αi)
for i = 1, ..., N+1.

Hence

(m*i, α*i)
= (mi, αi)

↓

(

ψ

ψμ

)

-1

for i = 1, ..., N+1.

Note that for every

(

ψ*

ψ*μ

)

∈H

and for i = 1, ..., N+1:

(mi, αi)↓

(

(

ψ

ψμ

)

-1

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

)

= (mi, αi)↓

(

ψ

ψμ

)

-1

↓

(

ψ*

ψ*μ

)

↓

(

ψ

ψμ

)

= (m*i, α*i)

↓

(

ψ*

ψ*μ

)

↓

(

ψ

ψμ

)

= (m*i, α*i)

↓

(

ψ

ψμ

)

= (mi, αi).

Hence

(

ψ

ψμ

)

-1

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

∈H for all

(

ψ*

ψ*μ

)

∈H

⇒

(

ψ

ψμ

)

∈NormG(H).
☐

23. LEMMA.There exists a Steiner system S(N+1,
2N, 6N ), where the points are the elements of the set
ZN]S3
and
the set of blocks is

Hence, there is at least one block, namely Fix↓(H),
that contains the points (m*1,
α*1),
..., (m*N+1,
α*N+1).
It remains to show that this is the unique block that contains the points
(m*1,
α*1),
..., (m*N+1,
α*N+1).
Suppose (m*1,
α*1),
..., (m*N+1,
α*N+1)
are contained in

Fix↓(H)↓

(

ψ*

ψ*μ

)

for some

(

ψ*

ψ*μ

)

∈G.

Then there exist points (m**1,
α**1),
..., (m**N+1,
α**N+1)
in Fix↓(H) such that

(m*i, α*i)
= (m**i, α**i)

↓

(

ψ*

ψ*μ

)

for i = 1, ..., N+1.

By lemma 22, there exists

(

ψ**

ψ**μ

)

∈NormG(H)

such that

(m**i, α**i)
= (mi, αi)

↓

(

ψ**

ψ**μ

)

for i = 1, ..., N+1.

Hence for i = 1, ..., N+1

(mi, αi)

↓

(

ψ

ψμ

)

= (m*i, α*i)

= (m**i, α**i)

↓

(

ψ*

ψ*μ

)

= (mi, αi)

↓

(

ψ**

ψ**μ

)

↓

(

ψ*

ψ*μ

)

⇒ (mi, αi)
= (mi, αi)↓

(

(

ψ**

ψ**μ

)

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

)

for i = 1, ..., N+1.

Then by lemma 17

(mi, αi)
= (mi, αi)↑

(

(

ψ**

ψ**μ

)

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

)

for i = 1, ..., N

and

(mN+1, αN+1)
= (mN+1, αN+1)↓

(

(

ψ**

ψ**μ

)

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

)

.

Hence

(

ψ**

ψ**μ

)

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

∈H.

Now H is a subgroup of NormG(H)

⇒

(

ψ**

ψ**μ

)

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

∈NormG(H)

⇒

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

∈

(

ψ**

ψ**μ

)

-1

NormG(H) = NormG(H)

⇒

(

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

)

H

(

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

)

-1

= H

⇒

(

ψ*

ψ*μ

)

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

(

ψ*

ψ*μ

)

-1

= H

⇒

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

=

(

ψ*

ψ*μ

)

-1

H

(

ψ*

ψ*μ

)

.

Now, using the first fact in the proof of lemma 22

Fix↓(H)↓

(

ψ*

ψ*μ

)

= Fix↓

(

(

ψ*

ψ*μ

)

-1

H

(

ψ*

ψ*μ

)

)

= Fix↓

(

(

ψ

ψμ

)

-1

H

(

ψ

ψμ

)

)

= Fix↓(H)↓

(

ψ

ψμ

)

.

This establishes the uniqueness of the block. ☐

24. THEOREM.Any map on the sphere can be properly coloured
by using at most four colours.

PROOF: Referring to section I, we have defined N to be the minimal
number of colours required to properly colour any map from the class of
all maps on the sphere. Based on the definition of N, we have selected
a specific map m(N) on the sphere which requires no
fewer than N colours to be properly coloured. Based on the definition
of the map
m(N) we have selected a proper colouring
of its regions using the N colours 0, 1, ..., N-1. Working
with the fixed number N, the fixed map
m(N),
and the fixed proper colouring of the regions of the map m(N),
lemma 23 has explicitly constructed a Steiner system S(N+1,
2N,6N). Now lemma 3 implies that N cannot
exceed four. ☐